Kerala SSLC Maths Model Question Paper 3 English Medium

Students can Download Kerala SSLC Maths Model Question Paper 3 English Medium, Kerala SSLC Maths Model Question Papers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala SSLC Maths Model Question Paper 3 English Medium

Time: 2½ Hours
Total Score: 80 Marks

Instructions

• Read each question carefully before writing the answer.
• Give explanations wherever necessary.
• First 15 minutes is Cool-off time. You may use the time to read the questions and plan your answers.
• No need to simplify irrationals like √2, √3, π etc., using approximations unless you are asked to do so.

Answer any three questions from 1 to 4. Each question carries 2 scores. (3 × 2 = 6)

Question 1.
The algebraic form of an arithmetic sequence is 6n + 8.
a) Find its first term.
b) What is the difference between its 100^m term and 98^m term?
Answer:
a) 6 × 1 + 8 = 14
b) 2 × 6 = 12

Question 2.
The centre of a circle is (0, 0), (4, 0) is a point on the circumference of the circle.
a) Find the radius of the circle.
b) Write the coordinates of another point on the circle.
Answer:
a) 4
b) (-4, 0)

Question 3.
In a place, there were 80 houses which as occupied. Due to recent floods \(\frac{3}{8}\) of houses were left.
a) Now how many houses were vacant?
b) What is the probability that a house is occupied?
Answer:
a) 80 × \(\frac{3}{8}\) = 30
b) 80 – 30 = 50
∴ \(\frac{50}{80}=\frac{5}{8}\)

Question 4.
Daily wages of some families are given below
350, 425, 280, 2300, 500
a) Find Mean
b) Find Median
Answer:
a) (350 + 425 + 280 + 2300 + 500)/5
= \(\frac{3855}{5}\)
= 771
b) 280, 350, 425, 500, 2300
∴ Median = 425

Write any 5 questions from 5 to 11. Each question carries 3 marks each. (5 × 3 = 15)

Question 5.
a) Draw a line of length √5 cm.
b) Draw a right triangle of base 6 cm and height √5 cm.
Answer:

Draw AB = 6 m.
Divide AB into AP, PB such that 5, 1.
Draw a circle such that AB as diameter.
Draw a perpendicular at P then extend B to R such that BR = 5 cm.
Then draw ΔPQR

Question 6.
P(x) = x² + ax + b
x – 2, x – 3 are the factors of P(x)
a) What will be the values of P(2), P(3).
b) Find the values of a, b.
Answer:
a) 2² + a × 2 + b = 2a + b + 4
3² + a × 3 + b = 9 + 3a + b = 3a + b + 9
b) 2a + b + 4 = 0 ⇒ 2a + b = -4 ……(1)
3a + b + 9 = 0 ⇒ 3a + b = -9 ………(2)
(1) – (2) ⇒ -a = 5 ⇒ a = -5
2 × – 5 + b = -4
⇒ b = -4 + 10 = 6

Question 7.
Area of a square pyramid is 360 cm². Area of one lateral face is 65 cm².
a) What is the slant height?
b) Find the height of the pyramid.
Answer:

Question 8.
(6, 4), (11, 11) are two points on a line.
a) Find the slope of this line.
b) Write two more points on this line.
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{11-4}{11-6}=\frac{7}{5}\)
If (11, 11) a point on this line, other points
(11 + 5, 11 + 7) = (16, 18)
(11 – 5, 11 – 7) = (6, 4)

Question 9.

In the fig: PA, PB are the tangents to the circle.
∠ACB = ∠APB
a) If ∠PAB = x°, then what will be ∠PBA
b) Then prove that ∆PAB is equilateral.
Answer:
a) x°
b) ∠ACB = x°
∠APB = x° = ∠PAB = ∠PBA
∆PAB is an equilateral triangle.

Question 10.

In the fig: OA = 8 unit
a) Find the coordinates of A.
b) Find the coordinates of B.
Answer:

Question 11.
A semicircle is rolled up to make a cone of radius 12 cm. Through the apex, and perpendicular to the base, It is cut into two pieces.
a) What will be the radius of the semicircle?
b) What is the slant height of the cone?
c) Which type of triangle is got after cutting it into two as mentioned above?
Answer:

b) l = 24 cm
c) Equilateral triangle

Answer any 7 questions from 12 to 21. Each question carries 4 scores. (7 × 4 = 28)

Question 12.
Sum of the first 5 terms of an AP is 320.
Sum of next 5 terms is 395.
a) Write the third term and the 8th term of this sequence.
b) What is the common difference?
c) Find the sum of the next 10 terms of this sequence.
Answer:

Question 13.
There are 16 mangoes in a basket of which 6 are unripe and others are ripe. Another basket contains 35 mangoes in which 13 are unripe and others are ripe. On the next day, 3 of the unripe mangoes in the first basket became ripe and 7 mangoes of the second basket became ripe.
a) What is the probability that we will get ripe from the first basket?
b) What is the probability that both being unripe if we took one mango from each on the first day?
c) What is the probability that we will get unripe mangoes from the second basket on the day ?
d) What is the probability that one being ripe and other, being unripe, if we took one from each on the second day?
Answer:

Question 14.
A boy of height 1.5 m is looking upon a tower at an elevation of 25°. The distance between the boy and the tower is 47 m.
a) Draw a rough figure.
b) Calculate the height of the tower.
[sin 25° = 0.4226, cos 25° = 0.9120, tan 25° = 0.4663]
Answer:
a)

b) tan 25 = \(\frac{A B}{B C}\)
0.4226 = \(\frac{A B}{47}\)
AB = 47 × 0.4226 = 19.86 m
Height of tower = 19.86 + 1.5 = 21.36 m

Question 15.
The table shows 53 children in a class sorted according to their heights

Height	No. of children
140 – 145	6
145 – 150	8
150 – 155	12
155 – 160	16
160 – 165	11

a) If the children are arranged in ascending order of height which student has median height?
b) Find the median height.
Answer:
Less than 145 – 6
Less than 150 – 14
Less than 155 – 26
Less than 160 – 42
Less than 165 – 53
a) 27th
b) 27th student
lies in the class 155-160 dividing 5 cm among 16 students

Question 16.
x² + y² – 6x – 8y + 9 = 0 is the equation of a circle.
a) Write the coordinates of the circle.
b) Find the radius.
Answer:
a) centre = (-g, -f)
\(=\left(-\frac{(-6)}{2},-\frac{(-8)}{2}\right)=(3,4)\)
b) (x² – 6x + 9) + (y² – 8y + 16) = -9 + 9 + 16
(x – 3)² + (y – 4)² = 42
Radius = 4

Question 17.
a) Draw a circle of radius 2 cm.
b) Draw a triangle whose two angles are 70°, 60° also its sides touches the circle.
Answer:

Question 18.
In figure AP, AQ, BC are the tangents to the circle. The perimeter of ΔABC is 70 cm. BP = 10 cm, CQ = 7 cm.

a) Find the length of BC.
b) Find all the sides of ΔABC
Answer:

Question 19.
One asks 5 days more to complete a job than the second one. If both of them works together the job will finish within 6 days.
a) If the number of days to finish the job for the second one is x, how many days will be needed for the first one.
b) From a second degree equation and find the no. of days required by each one to complete the job.
Answer:
a) x + 5
b) \(\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\)
\(\frac{x+5+x}{x(x+5)}=\frac{1}{6}\)
\(\frac{2 x+5}{x^{2}+5 x}=\frac{1}{6}\)
x² + 5x = 12x + 30
x² – 7x – 30 = 0
(x – 10) (x + 3) = 0
x = 10, -3
Time taken by both 10, 10 + 5 = 10, 15

Question 20.
In figure AB, AC is tangents to the circle ∠OBC = 30°. Find the following angles.

a) ∠OCB
b) ∠BOC
c) ∠A
d) If ΔPBC is an equilateral triangle find ∠PCAQ
Answer:
a) ∠OCB = 30°
b) ∠BOC = 120
c) ∠A = 60° (180 – 120)
d) ∠PCQ = 60°

Question 21.
P(x) = 4x² + 6x + k
a) If P(-2) = 0 then write a factor of P(x).
b) Find the value of k.
Answer:
a) x + 2
b) P(2) = 0
⇒ 4 × (-2)² + 6 × -2 + k = 0
⇒ 4 × 4 – 12 + k = 0
⇒ 16 – 12 + k = 0
⇒ 4 + k = 0
⇒ k = -4

Answer any 5 questions from 22 to 28. Each question carries 5 marks. (5 × 5 = 25)

Question 22.
A conical vessel of base radius 12 cm and height 36 cm contains water up to 9 cm high. After closing the lid the vessel is turned upside down as shown.

a) Find the volume of water in the first figure.
b) Find the volume of vacant space in the second figure.
c) Find the height of the vacant part of the vessel in the second figure.
Answer:

Question 23.
a) Write the arithmetic sequence between 100 and 1000 which can be divisible by 7.
b) What is the common difference.
c) What is the last term.
d) Form an algebric expression to find the sum of ‘n’ terms of this sequence.
Answer:
a) 105, 112, 119, …….
b) 7
c) 994
d) \(\frac{n}{2}\) (2f + (n – 1)d)

Question 24.
a) Draw a triangle whose sides are 6 cm, 8 cm and angle between them is 70°.
b) Draw a rectangle whose area is equal to the triangle.
c) Draw another rectangle with side 7 cm and equal area of the first rectangle.
Answer:

Question 25.

Cartres of the two circles lies on the same line.
AP = 4 cm, CD = 2.5 cm, BF = 7 cm.
a) Find all the lenghts of sides of ∆ABC
b) If the radius of small circle is 3 cm, find the area of ∆ABC
Answer:
a) AB = AP + BF = 4 + 7 = 11 cm
BC = BF + CD = 7 + 2.5 = 9.5 cm
AC = AP + CD = 4 + 2.5 = 6.5 cm
b) r = \(\frac{11+9.5+6.5}{2}=\frac{27}{2}=13.5\)
Area = 3r = 3 × 13.5 = 40.5 cm²

Question 26.
A(2, 2), B(7, 4), C(5, 6) are the vertices of quadrilateral ABCD. P, Q, R and S are the midpoints of AB, BC, CD, AD.
a) Find the coordinates of P, Q, R, S.
b) Prove that the perimeter of PQRS is AC + BD.
Answer:

b) PQRS is a parallelogram.
SR || AC. Also SR = ½ AC
PQ || AC. Also PQ = ½ AC
SP || BD, SP = ½ BD
QR || BD, QR = ½ BD
Perimeter of PQRS = PQ + SR + SP + QR
= ½ AC + ½ AC + ½ BD + ½ BD
= AC + BD

Question 27.

In the figure PQ is the tangent, AB is the chord. ∠BAQ = 54°, AB = 8 cm.
a) Find ∠AOB
b) Find the radius of the circle.
c) If AC = BC then find the perimeter of ∆ABC (sin 54 = 0.81, sin 63 = 0.89)
Answer:
a) ∠AOB = 2 × 54 = 108°

AC = 2 × 4.94 × 0.89 = 8.79 cm
BC = 8.79
Perimeter = 8.79 + 8.79 + 8 = 25.88 cm

Question 28.
When Many opened his social textbooks, he found the product of page numbers in front of him in 1122.
a) If the first-page number is ‘x’, what is the next page number?
b) Form a second-degree equation and find out the page numbers.
c) In other occasions, when he added 6 with the product of page numbers he got 2759. So which are the page numbers…?
Answer:
a) x + 1
b) x (x + 1) = 1122
x² + x – 1122 = 0
(x + 34) (x – 33) = 0
x = 33
Page numbers = 33, 34
c) x (x + 1) + 3 = 2759
x² + x + 3 = 2759
x² + x – 2756 = 0
(x + 53) (x – 52) = 0
x = 52
Page Numbers = 52, 53

Read the mathematical concept carefully and an¬swer the following question. (6 × 1 = 6)

Question 29.
In polygons line joining the opposite vertices are the diagonals. A triangle has no diagonal. A quadrilateral has two diagonal and a paragon five.
The number of diagonals of a polygon written in order is 0, 2, 5, ………..

This can be expressed in terms of their sides.

Like this, there is a beautiful relation between the number of sides and the number of diagonals. That can be drawn from one vertex of a polygon that is also written in order ………..
No. of sides : 3, 4, 5, …………
No. of diagonals from one vertex : 0, 1, 2, …………
Subtract 3 from the number of sides.
a) How many diagonals can be drawn from one vertex of a hexagon?
b) How many diagonals are there in a hexagon?
c) What is the number of diagonals of a polygon with ‘n’ sides?
d) How many sides are there for a polygon with 35 diagonals?
e) How many more diagonals are there in a polygon with 51 sides than that of a polygon with 50 sides?
f) How many more diagonals are there in a polygon with (n + 1) sides that of a polygon with ‘n’ sides?
Answer: